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How to replace string values with "XXXXX" in java?


By : fyzm7357
Date : July 29 2020, 02:00 PM
it fixes the issue I want to replace particular string values with "XXXX". The issue is the pattern is very dynamic and it won't have a fixed pattern in input data. , This may cover given both cases.
code :
String input = "https://internetbanking.abc.co.uk/personal/logon/login/?userId=Templ3108&password=Got1&reme";

// Mask UserId
System.out.println(input.replaceAll("userId=(\\w+)", "userId=XXXXX"));

// Mask Password
System.out.println(input.replaceAll("password=(\\w+)", "password=XXXXX"));
String maskUserNameAndPassword(String input) {
    return input.replaceAll("userId=(\\w+)", "userId=XXXXX")
                .replaceAll("password=(\\w+)", "password=XXXXX");
}


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Exception in thread "main" java.lang.NoClassDefFoundError: xxxxx (wrong name: xxxxx) error when running progra


By : Ignacio Hernandez
Date : March 29 2020, 07:55 AM
seems to work fine Based on the error message, I can see that Tester is inside a package com.gc.android.market.api
I would expect your console command to be something like

how to replace "?" with in a string with different array values using java/javascript


By : Olaf Tomalka
Date : March 29 2020, 07:55 AM
Hope this helps I want to replace the string "?" with a different array values in the order using javascript , One possible solution is this:
code :
var query = 'SELECT "x0"."NAME", "x0"."DOB", "x0"."SEX", "x0"."CATEGORY"' +
                    'FROM "PERSON" "x0"' +
                    'WHERE ((("x0"."NAME" <> ?) AND ("x0"."NAME" = ?)) AND 

("x0"."NAME" < ?))';

var names = ['borr', 'odin', 'thor'];

for (var i = 0; i < names.length; i++) {
    query = query.replace(/\?/, '"' + names[i] + '"');
}

console.log(query)

How do I change the value in "Xxxxx" format even if user inputs like this "xXXxX"? Java


By : anurag singla
Date : March 29 2020, 07:55 AM
To fix this issue I wrote a small utility for my self. It might help you. I was sure that my string would always have alphabets in it so did not care about any other characters. you might want to modify as per your needs.
code :
int length = "yourstring".length();/// get the length of the string
        String camelCase = removeCharacters.substring(0, 1).toUpperCase();// upper case the first alphabet
        camelCase = camelCase + "yourstring".substring(1, length).toLowerCase();// lowercase all other alphabets

postgres ERROR: insert or update on table "xxxxx" violates foreign key contrain "xxxxx"


By : Jack Tyson
Date : March 29 2020, 07:55 AM
help you fix your problem Did you read through the inheritance docs? Especially the 5.8.1. Caveats section?
http://www.postgresql.org/docs/9.0/static/ddl-inherit.html

The "replace" filter expects an array or "Traversable" as replace values, got "string"


By : Appolia
Date : March 29 2020, 07:55 AM
seems to work fine I had a similar issue - caused I think by an updated version of twig but not 100% sure. I suspect, that somewhere in your twig template (perhaps another not in your examples that is being extended?) you have a replace function like this:
{{something|replace('x','y')}}
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